Here's the question you clicked on:
zaphod
can anyone explain me, how to do this :)
@Callisto @ParthKohli @UnkleRhaukus @Hero
|dw:1351326687906:dw| As you can see, the total horizontal force is \[F_x=2F cos \theta \\ where\; cos \theta=\frac{3}{5}\]
|dw:1351326544993:dw| In X axis F=2T cos x where x=30/50=3/5[it is twice as OA and AC both components are same] so F=2*4*3/5=24/5=4.8N i.e (C)
But actually if i talk for exactness the information is wrong as elastic force is directly proportional to streching and it changes as we releases so integration is involved and lots more:)
so u mean none of a , b ,c & d.................:0
so its like this the resultant of both tension = force
Why have people stopped giving medals for the help they receive or the responses they like? @Aperogalics deserves a medal for his time and effort :s
ya this time u r correct HIS:) @rajathsbhat
sorry bro:( i was busy doing my work anyway u both deserve a medal. my apologizes again..
@zaphod i don't work 4 a medal i work 4 only giving knowledge and @rajathsbhat don't be sry it just happens mostly with me:P lolz
A medal is just a way to say "Your help is appreciated". And don't you agree that it feels good to have someone say that to you?
i really dont know what a medal means here as i join it just 2 days ago:)
@zaphod as the question is completed just close it:)